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 Page 1


JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Maths     Page | 15
Date : 2
nd
 September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Maths
Q.1 Let 
f : R R ?
be a function which satisfies f (x y) f (x) f (y) x, y R ? ? ? ? ? . If f(1) = 2
and 
(n 1)
k 1
g(n) f (k), n N
?
?
? ?
?
 then the value of n, for which g(n) = 20, is:
(1) 9 (2) 5 (3) 4 (4) 20
Sol. (2)
f(1) = 2 ; f(x+y) =f(x)+f(y)
x = y = 1 ? f(2) = 2+2 = 4
x = 2, y=1 ? f(3)=4+2 = 6
g(n)=f(1)+f(2)+.........+f(n-1)
= 2 + 4 + 6 + ........+ 2(n-1)
= 2 
?
(n-1)
= 2 
(n 1).n
2
?
= n
2
–n
Given g(n) = 20 ? n
2
 – n = 20
n
2
 – n–20 = 0
n = 5
Q.2 If the sum of first 11 terms of an A.P., a
1
, a
2
, a
3
, .... is 0(
1
a 0 ? ) then the sum of the A.P., ,
a
1
, a
3
, a
5
, ..., a
23
 is ka
1
, where k is equal to:
(1 ) 
121
10
?
(2) 
72
5
?
(3) 
72
5
(4) 
121
10
Sol. (2)
11
k
k 1
a
?
? =0 ? 11a + 55d = 0
a + 5d = 0
Now a
1
 +a
3
 + ....+ a
23
 = ka
1
12a + d (2+4+6+.....+22) = ka
12a + 2d. 66 = ka
12(a+11d) = ka
12
a
a 11 ka
5
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
12
11
1 k
5
? ?
? ?
? ?
? ?
k = —
72
5
Page 2


JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Maths     Page | 15
Date : 2
nd
 September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Maths
Q.1 Let 
f : R R ?
be a function which satisfies f (x y) f (x) f (y) x, y R ? ? ? ? ? . If f(1) = 2
and 
(n 1)
k 1
g(n) f (k), n N
?
?
? ?
?
 then the value of n, for which g(n) = 20, is:
(1) 9 (2) 5 (3) 4 (4) 20
Sol. (2)
f(1) = 2 ; f(x+y) =f(x)+f(y)
x = y = 1 ? f(2) = 2+2 = 4
x = 2, y=1 ? f(3)=4+2 = 6
g(n)=f(1)+f(2)+.........+f(n-1)
= 2 + 4 + 6 + ........+ 2(n-1)
= 2 
?
(n-1)
= 2 
(n 1).n
2
?
= n
2
–n
Given g(n) = 20 ? n
2
 – n = 20
n
2
 – n–20 = 0
n = 5
Q.2 If the sum of first 11 terms of an A.P., a
1
, a
2
, a
3
, .... is 0(
1
a 0 ? ) then the sum of the A.P., ,
a
1
, a
3
, a
5
, ..., a
23
 is ka
1
, where k is equal to:
(1 ) 
121
10
?
(2) 
72
5
?
(3) 
72
5
(4) 
121
10
Sol. (2)
11
k
k 1
a
?
? =0 ? 11a + 55d = 0
a + 5d = 0
Now a
1
 +a
3
 + ....+ a
23
 = ka
1
12a + d (2+4+6+.....+22) = ka
12a + 2d. 66 = ka
12(a+11d) = ka
12
a
a 11 ka
5
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
12
11
1 k
5
? ?
? ?
? ?
? ?
k = —
72
5
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Maths     Page | 16
Q.3 Let E
C
 denote the complement of an event E. Let E
1
, E
2
 and E
3
 be any pairwise independent
events with P(E
1
)>0 and 
1 2 3
P(E E E ) 0 ? ? ? . Then ? ?
C C
2 3 1
P E E / E ? is equal to:
(1) ? ? ? ?
C C
3 2
P E P E ? (2) ? ? ? ?
C
3 2
P E P E ?
(3) ? ? ? ?
C
3 2
P E P E ? (4) ? ? ? ?
C
2 3
P E P E ?
Sol. (3)
? ?
c c
2 3 1
P E E / E ? = 
c c
2 3 1
1
P(E E E )
P(E )
? ?
= 
1 1 2 1 3 1 2 3
1
P(E ) P(E E ) P(E E ) P(E E E )
P(E )
? ? ? ? ? ? ?
= 
1 1 2 1 3
1
P(E ) P(E ).P(E ) P(E ).P(E ) 0
P(E )
? ? ?
 = 1— P(E
2
) —P(E
3
)
= ? ?
c
3
P E — ? ?
2
P E
Q.4 If the equation 
4 4
cos sin 0 ? ? ? ? ? ?
 has real solutions for 
?
, then 
?
 lies in the interval:
(1) 
1 1
,
2 4
? ?
? ?
?
?
? ?
(2) 
1
1,
2
? ?
? ?
? ?
? ?
(3) 
3 5
,
2 4
? ?
? ?
? ?
? ?
(4) 
5
, 1
4
? ?
? ?
? ?
? ?
Sol. (2)
cos
4
? + sin
4
? + ? = 0
2
1
1 sin 2
2
? ?
? ? ? ? ?
? ?
? ?
2( ?+1) = sin
2
2 ?
0 < 2 ( ?+1) < 1
0 < ? + 1 < 
1
2
1
1
2
? ? ? ? ?
Q.5 The area (in sq. units) of an equilateral triangle inscribed in the parabola y
2
 = 8x, with
one of its vertices on the vertex of this parabola, is:
(1) 
128 3
(2) 
192 3
(3) 
64 3
(4) 
256 3
Sol. (2)
30
0
A
B
o
a
Page 3


JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Maths     Page | 15
Date : 2
nd
 September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Maths
Q.1 Let 
f : R R ?
be a function which satisfies f (x y) f (x) f (y) x, y R ? ? ? ? ? . If f(1) = 2
and 
(n 1)
k 1
g(n) f (k), n N
?
?
? ?
?
 then the value of n, for which g(n) = 20, is:
(1) 9 (2) 5 (3) 4 (4) 20
Sol. (2)
f(1) = 2 ; f(x+y) =f(x)+f(y)
x = y = 1 ? f(2) = 2+2 = 4
x = 2, y=1 ? f(3)=4+2 = 6
g(n)=f(1)+f(2)+.........+f(n-1)
= 2 + 4 + 6 + ........+ 2(n-1)
= 2 
?
(n-1)
= 2 
(n 1).n
2
?
= n
2
–n
Given g(n) = 20 ? n
2
 – n = 20
n
2
 – n–20 = 0
n = 5
Q.2 If the sum of first 11 terms of an A.P., a
1
, a
2
, a
3
, .... is 0(
1
a 0 ? ) then the sum of the A.P., ,
a
1
, a
3
, a
5
, ..., a
23
 is ka
1
, where k is equal to:
(1 ) 
121
10
?
(2) 
72
5
?
(3) 
72
5
(4) 
121
10
Sol. (2)
11
k
k 1
a
?
? =0 ? 11a + 55d = 0
a + 5d = 0
Now a
1
 +a
3
 + ....+ a
23
 = ka
1
12a + d (2+4+6+.....+22) = ka
12a + 2d. 66 = ka
12(a+11d) = ka
12
a
a 11 ka
5
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
12
11
1 k
5
? ?
? ?
? ?
? ?
k = —
72
5
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Maths     Page | 16
Q.3 Let E
C
 denote the complement of an event E. Let E
1
, E
2
 and E
3
 be any pairwise independent
events with P(E
1
)>0 and 
1 2 3
P(E E E ) 0 ? ? ? . Then ? ?
C C
2 3 1
P E E / E ? is equal to:
(1) ? ? ? ?
C C
3 2
P E P E ? (2) ? ? ? ?
C
3 2
P E P E ?
(3) ? ? ? ?
C
3 2
P E P E ? (4) ? ? ? ?
C
2 3
P E P E ?
Sol. (3)
? ?
c c
2 3 1
P E E / E ? = 
c c
2 3 1
1
P(E E E )
P(E )
? ?
= 
1 1 2 1 3 1 2 3
1
P(E ) P(E E ) P(E E ) P(E E E )
P(E )
? ? ? ? ? ? ?
= 
1 1 2 1 3
1
P(E ) P(E ).P(E ) P(E ).P(E ) 0
P(E )
? ? ?
 = 1— P(E
2
) —P(E
3
)
= ? ?
c
3
P E — ? ?
2
P E
Q.4 If the equation 
4 4
cos sin 0 ? ? ? ? ? ?
 has real solutions for 
?
, then 
?
 lies in the interval:
(1) 
1 1
,
2 4
? ?
? ?
?
?
? ?
(2) 
1
1,
2
? ?
? ?
? ?
? ?
(3) 
3 5
,
2 4
? ?
? ?
? ?
? ?
(4) 
5
, 1
4
? ?
? ?
? ?
? ?
Sol. (2)
cos
4
? + sin
4
? + ? = 0
2
1
1 sin 2
2
? ?
? ? ? ? ?
? ?
? ?
2( ?+1) = sin
2
2 ?
0 < 2 ( ?+1) < 1
0 < ? + 1 < 
1
2
1
1
2
? ? ? ? ?
Q.5 The area (in sq. units) of an equilateral triangle inscribed in the parabola y
2
 = 8x, with
one of its vertices on the vertex of this parabola, is:
(1) 
128 3
(2) 
192 3
(3) 
64 3
(4) 
256 3
Sol. (2)
30
0
A
B
o
a
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Maths     Page | 17
A : (a cos 30
0
, a sin 30
0
)
lies on parabola
2
a
4
=8.
a. 3
2
a 16 3 ?
Area of equilateral ? = 
3
4
a
2
? = 
3
4
. 16. 16. 3
? = 192
3
Q.6 The imaginary part of 
? ? ? ?
1/ 2 1/ 2
3 2 54 3 2 54 ? ? ? ? ? can be :
(1) 
6
(2) 
2 6 ?
(3) 6 (4) 
6 ?
Sol. (2)
? ?
1/2
3 2i 54 ? –
? ?
1/2
3 2i 54 ?
= 
? ?
1/2
2
9 6i 2.3i 6 ? ? –
? ?
1/2
2
9 6i 2.3i 6 ? ?
= ? ?
1/2
2
3 6i
? ?
?
? ?
? ?
- ? ?
1/2
2
3 6i
? ?
?
? ?
? ?
= ? ? ? ?
3 6i 3 6i ? ? ? ?
 = -2 6 i
Q.7 A plane passing through the point (3,1,1) contains two lines whose direction ratios are
1,–2,2 and 2,3, –1 respectively. If this plane also passes through the point ( ? ,–3,5),
then ? is equal to:
(1) –5 (2) 10 (3) 5 (4) –10
Sol. (3)
Required plane is
x 3 y 1 z 1
1 2 2
2 3 1
? ? ?
?
?
 = 0
? – 4(x – 3) + 5(y – 1) + 7(z – 1) = 0
? 4x – 5y – 7z = 0
( ? ?– 3, 5) lies on 4x – 5y – 7z = 0
? ? = 5
Q.8 Let A={X=(x, y, z)
T
: PX=0 and x
2
+y
2
+z
2
 = 1} ,where 
1 2 1
P 2 3 4
1 9 1
? ?
? ?
? ? ?
? ?
? ? ?
? ?
, then the set A:
(1) contains more than two elements (2) is a singleton.
(3) contains exactly two elements (4) is an empty set.
Page 4


JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Maths     Page | 15
Date : 2
nd
 September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Maths
Q.1 Let 
f : R R ?
be a function which satisfies f (x y) f (x) f (y) x, y R ? ? ? ? ? . If f(1) = 2
and 
(n 1)
k 1
g(n) f (k), n N
?
?
? ?
?
 then the value of n, for which g(n) = 20, is:
(1) 9 (2) 5 (3) 4 (4) 20
Sol. (2)
f(1) = 2 ; f(x+y) =f(x)+f(y)
x = y = 1 ? f(2) = 2+2 = 4
x = 2, y=1 ? f(3)=4+2 = 6
g(n)=f(1)+f(2)+.........+f(n-1)
= 2 + 4 + 6 + ........+ 2(n-1)
= 2 
?
(n-1)
= 2 
(n 1).n
2
?
= n
2
–n
Given g(n) = 20 ? n
2
 – n = 20
n
2
 – n–20 = 0
n = 5
Q.2 If the sum of first 11 terms of an A.P., a
1
, a
2
, a
3
, .... is 0(
1
a 0 ? ) then the sum of the A.P., ,
a
1
, a
3
, a
5
, ..., a
23
 is ka
1
, where k is equal to:
(1 ) 
121
10
?
(2) 
72
5
?
(3) 
72
5
(4) 
121
10
Sol. (2)
11
k
k 1
a
?
? =0 ? 11a + 55d = 0
a + 5d = 0
Now a
1
 +a
3
 + ....+ a
23
 = ka
1
12a + d (2+4+6+.....+22) = ka
12a + 2d. 66 = ka
12(a+11d) = ka
12
a
a 11 ka
5
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
12
11
1 k
5
? ?
? ?
? ?
? ?
k = —
72
5
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Maths     Page | 16
Q.3 Let E
C
 denote the complement of an event E. Let E
1
, E
2
 and E
3
 be any pairwise independent
events with P(E
1
)>0 and 
1 2 3
P(E E E ) 0 ? ? ? . Then ? ?
C C
2 3 1
P E E / E ? is equal to:
(1) ? ? ? ?
C C
3 2
P E P E ? (2) ? ? ? ?
C
3 2
P E P E ?
(3) ? ? ? ?
C
3 2
P E P E ? (4) ? ? ? ?
C
2 3
P E P E ?
Sol. (3)
? ?
c c
2 3 1
P E E / E ? = 
c c
2 3 1
1
P(E E E )
P(E )
? ?
= 
1 1 2 1 3 1 2 3
1
P(E ) P(E E ) P(E E ) P(E E E )
P(E )
? ? ? ? ? ? ?
= 
1 1 2 1 3
1
P(E ) P(E ).P(E ) P(E ).P(E ) 0
P(E )
? ? ?
 = 1— P(E
2
) —P(E
3
)
= ? ?
c
3
P E — ? ?
2
P E
Q.4 If the equation 
4 4
cos sin 0 ? ? ? ? ? ?
 has real solutions for 
?
, then 
?
 lies in the interval:
(1) 
1 1
,
2 4
? ?
? ?
?
?
? ?
(2) 
1
1,
2
? ?
? ?
? ?
? ?
(3) 
3 5
,
2 4
? ?
? ?
? ?
? ?
(4) 
5
, 1
4
? ?
? ?
? ?
? ?
Sol. (2)
cos
4
? + sin
4
? + ? = 0
2
1
1 sin 2
2
? ?
? ? ? ? ?
? ?
? ?
2( ?+1) = sin
2
2 ?
0 < 2 ( ?+1) < 1
0 < ? + 1 < 
1
2
1
1
2
? ? ? ? ?
Q.5 The area (in sq. units) of an equilateral triangle inscribed in the parabola y
2
 = 8x, with
one of its vertices on the vertex of this parabola, is:
(1) 
128 3
(2) 
192 3
(3) 
64 3
(4) 
256 3
Sol. (2)
30
0
A
B
o
a
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Maths     Page | 17
A : (a cos 30
0
, a sin 30
0
)
lies on parabola
2
a
4
=8.
a. 3
2
a 16 3 ?
Area of equilateral ? = 
3
4
a
2
? = 
3
4
. 16. 16. 3
? = 192
3
Q.6 The imaginary part of 
? ? ? ?
1/ 2 1/ 2
3 2 54 3 2 54 ? ? ? ? ? can be :
(1) 
6
(2) 
2 6 ?
(3) 6 (4) 
6 ?
Sol. (2)
? ?
1/2
3 2i 54 ? –
? ?
1/2
3 2i 54 ?
= 
? ?
1/2
2
9 6i 2.3i 6 ? ? –
? ?
1/2
2
9 6i 2.3i 6 ? ?
= ? ?
1/2
2
3 6i
? ?
?
? ?
? ?
- ? ?
1/2
2
3 6i
? ?
?
? ?
? ?
= ? ? ? ?
3 6i 3 6i ? ? ? ?
 = -2 6 i
Q.7 A plane passing through the point (3,1,1) contains two lines whose direction ratios are
1,–2,2 and 2,3, –1 respectively. If this plane also passes through the point ( ? ,–3,5),
then ? is equal to:
(1) –5 (2) 10 (3) 5 (4) –10
Sol. (3)
Required plane is
x 3 y 1 z 1
1 2 2
2 3 1
? ? ?
?
?
 = 0
? – 4(x – 3) + 5(y – 1) + 7(z – 1) = 0
? 4x – 5y – 7z = 0
( ? ?– 3, 5) lies on 4x – 5y – 7z = 0
? ? = 5
Q.8 Let A={X=(x, y, z)
T
: PX=0 and x
2
+y
2
+z
2
 = 1} ,where 
1 2 1
P 2 3 4
1 9 1
? ?
? ?
? ? ?
? ?
? ? ?
? ?
, then the set A:
(1) contains more than two elements (2) is a singleton.
(3) contains exactly two elements (4) is an empty set.
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Maths     Page | 18
Sol. (3)
Clearly |P| = 0
? PX = 0 has infinite solutions
The line concurrence passes through (0,0,0) which is centre of sphere x
2
 + y
2
 + z
2
 = 1
? ?Diameter will intersect at two points
? two solutions (exactly) exist
Q.9 The equation of the normal to the curve y=(1+x)
2y
 +cos
2
(sin
–1
x) at x=0 is:
(1) y+4x=2 (2) 2y+x=4 (3) x+4y=8 (4) y=4x+2
Sol. (3)
at x = 0 ? y = 1 + cos
2
(0)=2
p : (0,2)
y=(1+x)
2y
 +cos
2
(sin
–1
x)
y = (1 + x)
2y
 +  cos
2
? ?
–1 2
cos 1 – x
   = (1 + x)
2y
 +  
? ? ? ?
2
–1 2
cos cos 1– x
   = (1 + x)
2y
 +  
? ?
2
2
1– x
y = (1 + x)
2y
 + 1 – x
2
differentiating with respect to ‘x’
Now y’ = (1+x)
2y
? ?
? ? ?
? ?
?
? ?
2y
ln(1 x).2y' 2x
1 x
(0,2)
y'
= 4 –0
N
o
 : y – 2 = 
1
4
? (x–0)
N
o
 : 4y – 8 = -x
N
o
 : x 4y 8 ? ?
Q.10 Consider a region 
2
R {(x, y) R : ? ? 
2
x y 2x} ? ? . If a line y ? ? divides the area of
region R into two equal parts, then which of the following is true.?
(1) 
3 2
6 16 0 ? ? ? ? ?
(2) 
2 3/ 2
3 8 8 0 ? ? ? ? ?
(3) 
3 3/ 2
6 16 ? ? ? ?
(4) 
2
3 8 8 0 ? ? ? ? ?
Sol. (2)
x y 2x
2
< < 
y=x
2
y=2x
O 2
Area = 
? ?
2
2
0
2x x dx ?
?
 = x
2
 –
2
3
0
x
3
 = 
4
3
Page 5


JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Maths     Page | 15
Date : 2
nd
 September 2020
Time : 02 : 00 pm - 05 : 00 pm
Subject : Maths
Q.1 Let 
f : R R ?
be a function which satisfies f (x y) f (x) f (y) x, y R ? ? ? ? ? . If f(1) = 2
and 
(n 1)
k 1
g(n) f (k), n N
?
?
? ?
?
 then the value of n, for which g(n) = 20, is:
(1) 9 (2) 5 (3) 4 (4) 20
Sol. (2)
f(1) = 2 ; f(x+y) =f(x)+f(y)
x = y = 1 ? f(2) = 2+2 = 4
x = 2, y=1 ? f(3)=4+2 = 6
g(n)=f(1)+f(2)+.........+f(n-1)
= 2 + 4 + 6 + ........+ 2(n-1)
= 2 
?
(n-1)
= 2 
(n 1).n
2
?
= n
2
–n
Given g(n) = 20 ? n
2
 – n = 20
n
2
 – n–20 = 0
n = 5
Q.2 If the sum of first 11 terms of an A.P., a
1
, a
2
, a
3
, .... is 0(
1
a 0 ? ) then the sum of the A.P., ,
a
1
, a
3
, a
5
, ..., a
23
 is ka
1
, where k is equal to:
(1 ) 
121
10
?
(2) 
72
5
?
(3) 
72
5
(4) 
121
10
Sol. (2)
11
k
k 1
a
?
? =0 ? 11a + 55d = 0
a + 5d = 0
Now a
1
 +a
3
 + ....+ a
23
 = ka
1
12a + d (2+4+6+.....+22) = ka
12a + 2d. 66 = ka
12(a+11d) = ka
12
a
a 11 ka
5
? ? ? ?
? ? ?
? ? ? ?
? ? ? ?
12
11
1 k
5
? ?
? ?
? ?
? ?
k = —
72
5
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Maths     Page | 16
Q.3 Let E
C
 denote the complement of an event E. Let E
1
, E
2
 and E
3
 be any pairwise independent
events with P(E
1
)>0 and 
1 2 3
P(E E E ) 0 ? ? ? . Then ? ?
C C
2 3 1
P E E / E ? is equal to:
(1) ? ? ? ?
C C
3 2
P E P E ? (2) ? ? ? ?
C
3 2
P E P E ?
(3) ? ? ? ?
C
3 2
P E P E ? (4) ? ? ? ?
C
2 3
P E P E ?
Sol. (3)
? ?
c c
2 3 1
P E E / E ? = 
c c
2 3 1
1
P(E E E )
P(E )
? ?
= 
1 1 2 1 3 1 2 3
1
P(E ) P(E E ) P(E E ) P(E E E )
P(E )
? ? ? ? ? ? ?
= 
1 1 2 1 3
1
P(E ) P(E ).P(E ) P(E ).P(E ) 0
P(E )
? ? ?
 = 1— P(E
2
) —P(E
3
)
= ? ?
c
3
P E — ? ?
2
P E
Q.4 If the equation 
4 4
cos sin 0 ? ? ? ? ? ?
 has real solutions for 
?
, then 
?
 lies in the interval:
(1) 
1 1
,
2 4
? ?
? ?
?
?
? ?
(2) 
1
1,
2
? ?
? ?
? ?
? ?
(3) 
3 5
,
2 4
? ?
? ?
? ?
? ?
(4) 
5
, 1
4
? ?
? ?
? ?
? ?
Sol. (2)
cos
4
? + sin
4
? + ? = 0
2
1
1 sin 2
2
? ?
? ? ? ? ?
? ?
? ?
2( ?+1) = sin
2
2 ?
0 < 2 ( ?+1) < 1
0 < ? + 1 < 
1
2
1
1
2
? ? ? ? ?
Q.5 The area (in sq. units) of an equilateral triangle inscribed in the parabola y
2
 = 8x, with
one of its vertices on the vertex of this parabola, is:
(1) 
128 3
(2) 
192 3
(3) 
64 3
(4) 
256 3
Sol. (2)
30
0
A
B
o
a
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Maths     Page | 17
A : (a cos 30
0
, a sin 30
0
)
lies on parabola
2
a
4
=8.
a. 3
2
a 16 3 ?
Area of equilateral ? = 
3
4
a
2
? = 
3
4
. 16. 16. 3
? = 192
3
Q.6 The imaginary part of 
? ? ? ?
1/ 2 1/ 2
3 2 54 3 2 54 ? ? ? ? ? can be :
(1) 
6
(2) 
2 6 ?
(3) 6 (4) 
6 ?
Sol. (2)
? ?
1/2
3 2i 54 ? –
? ?
1/2
3 2i 54 ?
= 
? ?
1/2
2
9 6i 2.3i 6 ? ? –
? ?
1/2
2
9 6i 2.3i 6 ? ?
= ? ?
1/2
2
3 6i
? ?
?
? ?
? ?
- ? ?
1/2
2
3 6i
? ?
?
? ?
? ?
= ? ? ? ?
3 6i 3 6i ? ? ? ?
 = -2 6 i
Q.7 A plane passing through the point (3,1,1) contains two lines whose direction ratios are
1,–2,2 and 2,3, –1 respectively. If this plane also passes through the point ( ? ,–3,5),
then ? is equal to:
(1) –5 (2) 10 (3) 5 (4) –10
Sol. (3)
Required plane is
x 3 y 1 z 1
1 2 2
2 3 1
? ? ?
?
?
 = 0
? – 4(x – 3) + 5(y – 1) + 7(z – 1) = 0
? 4x – 5y – 7z = 0
( ? ?– 3, 5) lies on 4x – 5y – 7z = 0
? ? = 5
Q.8 Let A={X=(x, y, z)
T
: PX=0 and x
2
+y
2
+z
2
 = 1} ,where 
1 2 1
P 2 3 4
1 9 1
? ?
? ?
? ? ?
? ?
? ? ?
? ?
, then the set A:
(1) contains more than two elements (2) is a singleton.
(3) contains exactly two elements (4) is an empty set.
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Maths     Page | 18
Sol. (3)
Clearly |P| = 0
? PX = 0 has infinite solutions
The line concurrence passes through (0,0,0) which is centre of sphere x
2
 + y
2
 + z
2
 = 1
? ?Diameter will intersect at two points
? two solutions (exactly) exist
Q.9 The equation of the normal to the curve y=(1+x)
2y
 +cos
2
(sin
–1
x) at x=0 is:
(1) y+4x=2 (2) 2y+x=4 (3) x+4y=8 (4) y=4x+2
Sol. (3)
at x = 0 ? y = 1 + cos
2
(0)=2
p : (0,2)
y=(1+x)
2y
 +cos
2
(sin
–1
x)
y = (1 + x)
2y
 +  cos
2
? ?
–1 2
cos 1 – x
   = (1 + x)
2y
 +  
? ? ? ?
2
–1 2
cos cos 1– x
   = (1 + x)
2y
 +  
? ?
2
2
1– x
y = (1 + x)
2y
 + 1 – x
2
differentiating with respect to ‘x’
Now y’ = (1+x)
2y
? ?
? ? ?
? ?
?
? ?
2y
ln(1 x).2y' 2x
1 x
(0,2)
y'
= 4 –0
N
o
 : y – 2 = 
1
4
? (x–0)
N
o
 : 4y – 8 = -x
N
o
 : x 4y 8 ? ?
Q.10 Consider a region 
2
R {(x, y) R : ? ? 
2
x y 2x} ? ? . If a line y ? ? divides the area of
region R into two equal parts, then which of the following is true.?
(1) 
3 2
6 16 0 ? ? ? ? ?
(2) 
2 3/ 2
3 8 8 0 ? ? ? ? ?
(3) 
3 3/ 2
6 16 ? ? ? ?
(4) 
2
3 8 8 0 ? ? ? ? ?
Sol. (2)
x y 2x
2
< < 
y=x
2
y=2x
O 2
Area = 
? ?
2
2
0
2x x dx ?
?
 = x
2
 –
2
3
0
x
3
 = 
4
3
JEE Main 2020 Paper
         2
nd
 September 2020 | (Shift-2), Maths     Page | 19
? 
0
y 2
y dy
2 3
?
? ?
? ?
? ?
? ?
?
? ?
3/2
2
y
3
 – 
2
0
y
4
?
 = 
2
3
? ?
3/2
8. ? – 3 ?
2
 = 8
Q.11 Let f : ( 1, ) R ? ? ? be defined by f(0)=1 and 
e
1
f (x) log (1 x), x 0
x
? ? ?
. Then the function
f:
(1) increases in (–1,
?
)
(2) decreases in (–1,0) and increases in (0,
?
)
(3) increases in (–1,0) and decreases in (0,
?
)
(4) decreases in (–1,
?
).
Sol. (4)
f(x)=
1
n(1 x)
x
? ?
f
’
 = 
2
1
x In(1 x)
1 x
x
? ? ?
?
f
’
 = 
2
1
1 ln(1 x)
1 x
x
? ? ?
?
f' 0 x ( 1, ) ? ? ? ? ?
Q.12 Which of the following is a tautology?
(1) (p ?q)
?
(q ?p) (2) (~p)
?
(p
?
q) ?q
(3) (q ?p)
?
~(p ?q) (4) (~q)
?
(p
?
q) ?q
Sol. (2)
? ? ? ? p q ~ p pvq ~ p p q ~ p p q q
T T F T F T
T F F T F T
F T T T T T
F F T F F T
? ? ? ? ?
Q.13 Let f(x) be a quadratic polynomial such that f(–1)+f(2)=0. If one of the roots of f(x)=0 is
3, then its other roots lies in:
(1) (0,1) (2) (1,3) (3) (–1,0) (4) (–3,–1)
Sol. (3)
Let f(x) = a (x-3) (x- ?)
f(-1)+f(2)=0
a[(–1–3) (–1- ?)+(2-3)(2– ?)]=0
a[4+4 ?-2+ ?]=0
5 ?+2 = 0
2
5
? ? ?
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